# An object moving at speed 40cm/sec towards a concave mirror of focal kenght 20cm.When the object is at distance 60cm,the speed of image is

Piyush Kumar Behera
436 Points
3 years ago
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
Differentiating the expression w.r.t. time
$0=\frac{1}{v^{2}}\frac{du}{dt}+\frac{1}{u^{2}}\frac{dv}{dt}$
$\frac{dv}{dt}=-\frac{v^{2}}{u^{2}}\frac{du}{dt}$

Just find the final position of the image by using the mirror formula...
$\frac{1}{-20}=\frac{1}{-60}+\frac{1}{v}$

Solving we get
v = – 30 cm

Now using this formula we get
$\frac{dv}{dt}=-\frac{v^{2}}{u^{2}}\frac{du}{dt}$
$\frac{dv}{dt}=-\frac{(-30)^{2}}{(-60)^{2}} *40$

$\frac{dv}{dt}=-\frac{1}{4}*40$
$\frac{dv}{dt}=-10 cm/sec$

So the velocity of the image is 10 cm/sec opposite to the direction of the velocity of object.

Piyush Kumar Behera